常用等价无穷小

当 $x \rightarrow 0$ 时

$sin\,x \sim x$

$tan\,x \sim x$

$arcsin\,x \sim x$

$arctan\,x \sim x$

$ln(1+x) \sim x$

$e^x -1 \sim x$

$(1+x)^\alpha \sim \alpha x$

$log_a(1+x) \sim \frac{x}{ln \,a}$

不定积分公式

$\int \frac{1}{x} \,dx = ln\mid x \mid +C $

$\int tan\,x \,dx = -ln\mid cos\,x \mid +C$

$\int sec\,x \,dx=-ln \mid sec\,x + tan\,x \mid + C$

${\large \int \frac{dx}{\sqrt{a^2-x^2}} } = arcsin\,\frac{x}{a} + C $

${\large \int \frac{dx}{a^2+x^2} }=\frac1{a}arctan\,\frac{x}{a} +C$

${\large \int \frac{dx}{x^2-a^2} = \frac{1}{2a}ln \mid \frac{x-a}{x+a} \mid } +C$

${\large \int \frac{dx}{\sqrt{x^2+a^2}} } = ln(x+\sqrt{x^2+a^2})+C$

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${\large \int \frac{1}{2\sqrt{x}}dx = \sqrt{x}}$

${\large \int \frac{1}{\sqrt{x}} = 2\sqrt{x}}$

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${\large \int \frac{1}{x^2}dx = -\frac{1}{x}}$

${\large \int -\frac{1}{x^2}dx = \frac{1}{x}}$

常用泰勒展开

$(1+x)^\alpha = 1+ \displaystyle\sum^{\infty}_{n=1}\frac{\alpha(\alpha-1)···(\alpha-n+1)}{n!}x^n $

$= 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!}x^2+o(x^2)$

$tan\,x = x + \frac13x^3+\frac{2}{15}x^5+o(n^5)$

$sin\,x = x-\frac{1}{3!}x^3 +\frac{1}{5!}x^5+o(x^5)$

$cos\,x=1-\frac{1}{2!}x^2 + \frac{1}{4!}x^4 + o(x^4)$

$ln\,(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 + o(x^3)$

$\frac{1}{1-x} = 1 + x + x^2 + x^3 + o(x^3)$

$\frac{1}{1+x} = 1-x+x^2-x^3 + o(x^3)$

$arcsin\,x= x + \frac{1}{6}x^3 + \frac{3}{40}x^5 + \frac{5}{112}x^7 + o(x^7)$

$arccos\,x= \frac{\pi}{2} - arcsin\,x$

$arctan\,x = x - \frac{1}{3}x^3 + \frac{1}{5}x^5 + o(x^5)$

$e^x = 1 + \frac{1}{1!}x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + o(x^3)$

无穷大处的泰勒展开

$\displaystyle\lim_{x\to+\infty}\sqrt{x^2+1}=\displaystyle\lim_{x\to+\infty}x\sqrt{1+\frac1{x^2}}$

自然数的前n项的m次方和

$1^2 + 2^2 + …+n^2 = \frac16n(n+1)(2n+1)$

利用 $(n+1)^3-n^3=3n^2+3n+1$ 推导

$1^3 + 2^3 + … + n^3 = \frac14n^2(n+1)^2$

其他公式

$arcsin\,x+arccos\,x=\frac{\pi}{2}$

$arctan\,x+arctan\,\frac1x=\frac{\pi}{2}\quad ,x>0$

$arctan\,x+arctan\,\frac1x=-\frac{\pi}{2}\quad ,x<0$

均值不等式

调和平均数，几何平均数，算术平均数，平方平均数。

$H_n\leq G_n\leq A_n\leq Q_n$

$H_n=\displaystyle\frac{n}{\displaystyle\sum_{i=1}^{n}\frac{1}{x_i}}=\frac{n}{\displaystyle\frac1{x_1}+\frac{1}{x_2}+···+\frac{1}{x_n}}$

$G_n=\sqrt[n]{\displaystyle\prod_{i=1}^{n}x_i}=\sqrt[n]{\displaystyle x_1x_2···x_n}$

$A_n=\displaystyle\frac{\displaystyle\sum_{i=1}^{n}x_i}{n}=\frac{x_1+x_2+···+x_n}{n}$

$Q_n=\sqrt{\frac{\displaystyle\sum_{i=1}^{n}x^2_i}{\displaystyle n}}=\sqrt{\displaystyle\frac{x_1^2+x_2^2+···+x_n^2}{\displaystyle n}}$